The splitting lemma says:
Given a short exact sequence with maps $q$ and $r$:
$0 \rightarrow A \overset{q}{\rightarrow} B \overset{r}{\rightarrow} C \rightarrow 0$
then the following are equivalent:
- …
- there exists a map $u : C \rightarrow B$ such that $r \circ u = \mathrm{id}_C$
- $B \cong A \oplus C$
Now I figured, since $r(B) = \ker 0$, $r$ is surjective. Hence, for every $c \in C$ we have some $b \in B$ such that $r(b) = c$. Simply set $u(c) = b$ for one of these $b$s and you have your map $u$.
However, it turns out that this construction assumes the axiom of choice; it chooses one element from each of an infinite number of sets. So my question is: assuming the axiom of choice, is condition (2) always satisfied? Because this would imply that (3) holds for any such short exact sequence. Or am I making some mistake here?
Best Answer
I assume you are working in some fixed abelian category $\mathcal{A}$.
It is not true in general that every short exact sequence in $\mathcal{A}$ will split. The problem is that although you can pick a preimage for every 'element' $c\in C$ there is no guarantee that you can assemble this into a morphism in $\mathcal{A}$. It is true in the category of sets that every surjection splits if one assumes the axiom of choice but this is only a set map.
For instance in the category of abelian groups $$0 \to \mathbb{Z}/2\mathbb{Z} \stackrel{2}{\to} \mathbb{Z}/4\mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0$$ is exact but $\mathbb{Z}/4\mathbb{Z}$ is indecomposable.
Also if you are not in an abelian category it is not necessarily true that exact sequences display this symmetry. See for example this question where the notion is considered in the category of groups.